Induced maps of an automorphism of a curve on the tangent ot its jacobian and on its differential forms

Question

Let $C$ be a smooth projective curve of genus $g$ over a field $k$ and $J$ be its jacobian (defined over $k$). Let $\sigma: C \rightarrow C$ be a $k$-automorphsm of $C$. This automorphism $\sigma$ induces an automorphism (as abelian variety) on $J$, and hence on the tangent space $T$ at the origin $0_J$ of $J$. We know $\mathrm{dim}_k T = \mathrm{dim} \ J = g$. On the other hand, we can consider the sheaf of differential $\Omega_C$ on C. We know $\mathrm{dim}_k H^{0}(C, \Omega) = g$ and $\sigma$ induce an automorphism on $H^{0}(C, \Omega)$.

My questions is:

The induced maps of $\sigma$ on "the tangent $T$ of $J$" and on "the differential forms of $C$" are (canonical) related? (Over $\mathbb{C}$ only or true in general?)

In a book, the author mentioned that these two maps on vector spaces are the same without giving the reason. I am OK with this fact, but I just don't know how to see this. I would like to know if the identification is canoncial also?

Answer 1

Over any field, $T_0 J$ is canonically isomorphic to $H^1(C, \mathcal{O})$ and hence, by Serre duality, is dual to $H^0(C, \Omega^1)$. Seeing that $T_0 J \cong H^1(C, \mathcal{O})$ is a good example of using the functorial description of $J$.

Let $D = \mathrm{Spec}\ k[\epsilon]/\epsilon^2$. Then $T_0 J$ is the set of maps $D$ to $J$ such that the composite $\mathrm{Spec}\ k \to D \to J$ lands on $0$. For any $k$-scheme $S$, $\mathrm{Hom}(S, J)$ is the group of isomorphism classes of line bundles on $C \times S$, modulo line bundles pulled back from $S$. In the case of $D$, all line bundles on $D$ are trivial, so $\mathrm{Hom}(D, J)$ is isomorphism classes of line bundles on $D \times C$. The homs which send $\mathrm{Spec} \ k$ to $0$ are those where the line bundle on $(\mathrm{Spec} \ k) \times C$ is trivial.

Let $U = \mathrm{Spec} \ A$ be an open affine of $C$, so $D \times U$ is an open affine of $D \times C$. Note that $D \times U \cong \mathrm{Spec} A[\epsilon]/\epsilon^2$. Then we have a short exact sequence $$0 \to A_{+} \to (A[\epsilon]/\epsilon^2)^{\ast} \to A^{\ast} \to 0.$$ Here $B^{\ast}$ is the unit group of the ring $B$, and $A_{+}$ is $A$ considered as an additive group. The first map is $a \mapsto 1+a \epsilon$. This turns into a short exact sequence of sheaves on $C$: $$0 \to \mathcal{O}_C \to \mathcal{O}_{D \times C}^{\ast} \to \mathcal{O}_C^{\ast} \to 0.$$ (Note that $C$ and $D \times C$ are the same underlying topological space, so all of these are sheaves on the same space.)

So we have $H^1(C, \mathcal{O}) \to \mathrm{Pic}(D \times C) \to \mathrm{Pic}(C)$. We argued before that $T_0(J)$ is the kernel of the second map, and a little more work shows that the first map is injective, so $T_0(J) \cong H^1(C, \mathcal{O})$ as desired. "$\square$"

The end of proof symbol is in quotes, because one should really check that this isomorphism is compatible with the vector space structures and work out how it relates to endomorphisms of $C$, but this is the idea.

Answer 2

The two maps are naturally one the "dual" of the other. Let me give an immediate proof when $k=\mathbf{C}$.

Over the complex numbers one has $$J(C)=H^0(C, \Omega^1_C)^*/H_1(C, \mathbf{Z}),$$ hence $T_0J(C)=H^0(C, \Omega^1_C)^*$.

Therefore the automorphism $\sigma \colon C \to C$ induces an automorphism $\bar{\sigma} \colon H^0(C, \Omega^1_C)^* \to H^0(C, \Omega^1_C)^*$.

Dualizing it, we obtain the automorphism $\bar{\sigma}^* \colon H^0(C, \Omega^1_C) \to H^0(C, \Omega^1_C).$